Velocity of a ball?

Question

If a ball is thrown vertically upward from the roof of a 64 ft building with a velocity of 32 ft/sec it's height after t seconds is
s(t)=64+32t-16t^2

What is the velocity of the ball when it hits the ground (0)??

Answer 1

V^2 = 2gh or h= V^2/2g
Therefore the total height is 64 + 32^2/64 = 80 feet.

V = (2gh)^0.5 = (2*32*80)^0.5 = 71.6 feet/sec

Answer 2

When the reference point is from the ground, because you the 64 added to the original equation was accounting for the fact it was thrown from the top of a building. So when it hits the ground, you want to know the time. So set this equation to 0 (because you are looking for t when s=0) and solve:
-16t^2 + 32t + 64 = 0
-t^2 + 2t + 4 = 0
[-b +- root(b^2 - 4ac)]/2a
[-2 +- root(4 - 4(-1)(4))]/ 2(-1)
1 +- root(20)/2
1 +- 2root(5)/2
1 +- root(5)

Two answers, 1+root(5) and 1-root(5), but time can't negative, so use 1+root(5).

This is the time the ball hits the ground. To figure out velocity, use the equation v = v0 + at
v = 32 ft/sec + (-16 ft/s^2)(1+root(5))
v = - 19.78 ft/sec

So it hits the ground at 19.78 ft/sec

Answer 3

Conservation of momentum (linear:P and angular:L) P1 = 0.01 * 200 = 2 kgm/s P1 = L2 + P2 assuming a hundred% inelastic collision and negilgable mass income. could desire to locate rotational inertia of ball per geometry assuming homogeneous density. i've got have been given self concept it particularly is (2m/5)*R^2 detect instantaneous L2 after result. Calculate instantaneous P2 after result word acceleration imparted by using floor friction untill angular velocity suits translational velocity for not extra slippage. word: there will be a specfic result height the region the angular velocity will thoroughly negate the translational velocity after frictional acceleration is comprehensive and the ball could be left at loosen up. per possibility it particularly is R/2???

Answer 4

When throwing from the top of the building:-
v² = u² - 2 g s
0 = 32² - 20 s
s = 32² / 20
s = 51.2 ft
Ball then drops a distance of 51.2 + 64 ft
d = 115.2 ft
Velocity when hitting ground is Vg , say
(Vg) ² = u² + 2 g (115.2)
(Vg) ² = 0 + 2304
Vg = 48 ft / sec